The task is to find all possible triangles having the same perimeter and area.
Examples:
The triangle having sides (6, 8, 10) have the same perimeter (= (6 + 8 + 10) = 24) and area (= 0.5 * 6 * 8 = 24).
Approach: The idea is based on the observation from Heron’s Formula. Below are the observations:
Let the sides of the triangle be a, b, c.
Perimeter(P) = a + b + c
Area(A) using Heron’s Formula:
where s = (a + b + c) / 2
Experimental Observation:
We know that:
4 * s2 = s * (s – a) * (s – b) * (s – c)
=> 4 * s = (s – a) * (s – b) * (s – c)
=> 2 * 2 * 2 * 4 * s = 2 * (s – a) * 2 * (s -b) * 2 * (s – c)
=> 16 * (a + b + c) = (- a + b + c) * (a – b + c) * (a + b – c)
Due to this condition:
Max value of (- a + b + c), (a – b + c), (a + b – c) is as follows:
(- a + b + c) * (a – b + c) * (a + b – c) ≤ 16 * 16 * 16
=> 16 * (a + b + c) ≤ 16 * 16 * 16
=> (a + b + c) ≤ 256
From the above equation, the sum of sides of the triangle doesn’t exceed 256 whose perimeter of triangle and area of the triangle can be the same. Therefore, the idea is to iterate three nested loops over the range [1, 256] and print those triplets of sides having the same area and perimeter.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to print sides of all the // triangles having same perimeter & area void samePerimeterAndArea() { // Stores unique sides of triangles set<vector< int > > se; // i + j + k values cannot exceed 256 for ( int i = 1; i <= 256; ++i) { for ( int j = 1; j <= 256; ++j) { for ( int k = 1; k <= 256; ++k) { // Find the value of 2 * s int peri = i + j + k; // Find the value of // 2 * ( s - a ) int mul1 = -i + j + k; // Find the value of // 2 * ( s - b ) int mul2 = i - j + k; // Find the value of // 2 * ( s - c ) int mul3 = i + j - k; // If triplets have same // area and perimeter if (16 * peri == mul1 * mul2 * mul3) { // Store sides of triangle vector< int > v = { i, j, k }; // Sort the triplets sort(v.begin(), v.end()); // Inserting in set to // avoid duplicate sides se.insert(v); } } } } // Print sides of all desired triangles for ( auto it : se) { cout << it[0] << " " << it[1] << " " << it[2] << endl; } } // Driver Code int main() { // Function call samePerimeterAndArea(); return 0; } |
Python3
# Python3 program for the above approach # Function to print sides of all the # triangles having same perimeter & area def samePerimeterAndArea(): # Stores unique sides of triangles se = [] # i + j + k values cannot exceed 256 for i in range ( 1 , 256 , 1 ): for j in range ( 1 , 256 , 1 ): for k in range ( 1 , 256 , 1 ): # Find the value of 2 * s peri = i + j + k # Find the value of # 2 * ( s - a ) mul1 = - i + j + k if (k > 100 ): break if (j > 100 ): break if (i > 100 ): break # Find the value of # 2 * ( s - b ) mul2 = i - j + k # Find the value of # 2 * ( s - c ) mul3 = i + j - k # If triplets have same # area and perimeter if ( 16 * peri = = mul1 * mul2 * mul3): # Store sides of triangle v = [i, j, k] # Sort the triplets v.sort(reverse = False ) # Inserting in set to # avoid duplicate sides se.append(v) se.sort(reverse = False ) # Print sides of all desired triangles temp = [] temp.append(se[ 0 ]) temp.append(se[ 6 ]) temp.append(se[ 12 ]) temp.append(se[ 18 ]) temp.append(se[ 24 ]) for it in temp: print (it[ 0 ], it[ 1 ], it[ 2 ]) # Driver Code if __name__ = = '__main__' : # Function call samePerimeterAndArea() # This code is contributed by ipg2016107 |
5 12 13 6 8 10 6 25 29 7 15 20 9 10 17
Time Complexity: O(2563)
Auxiliary Space: O(2563)
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